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Example 17.8

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\n(17.48)Distinctintegersβˆ—nβˆ—andβˆ—mβˆ—havebeenintroducedtohandletheproductofthetwoseriessummations.Usingthesamereasoningasabove,weget>βˆ—Fβˆ—rms<sup>2</sup>=βˆ—aβˆ—<sup>0</sup>2<sup>+</sup>__12βˆ‘βˆ—nβˆ—=1βˆžβˆ—Anβˆ—2or\n(17.48) Distinct integers *n* and *m* have been introduced to handle the product of the tw o series summations. Using the same reasoning as abo ve, we get > *F*rms <sup>2</sup> = *a*<sup>0</sup> 2 <sup>+</sup>\_\_1 2 βˆ‘ *n*=1 ∞ *A n* 2 or

F_{\rm rms} = \sqrt{a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} A_n^2}

(17.49)IntermsofFouriercoefficientsβˆ—anβˆ—andβˆ—bnβˆ—,Eq.(17.49)maybewrittenascoefficients (17.49) In terms of Fourier coefficients *an* and *bn*, Eq. (17.49) may be written as coefficients

a_n

and $b_n$ , Eq. (17.49) may be written as \n

F_{\text{rms}} = \sqrt{a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} (a_n^2 + b_n^2)}

\n(17.50)Ifβˆ—fβˆ—(βˆ—tβˆ—)isthecurrentthrougharesistorβˆ—Rβˆ—,thenthepowerdissipatedintheresistoris\n(17.50) If *f*(*t*) is the current through a resistor *R*, then the power dissipated in the resistor is

P = RF_{\rm rms}^2 \tag{17.51}

Orifβˆ—fβˆ—(βˆ—tβˆ—)isthevoltageacrossaresistorβˆ—Rβˆ—,thepowerdissipatedintheresistoris Or if *f*(*t*) is the voltage across a resistor *R*, the power dissipated in the resistor is

P = \frac{F_{\text{rms}}^2}{R}

(17.52)Onecanavoidspecifyingthenatureofthesignalbychoosinga1βˆ’Ξ©resistance.Thepowerdissipatedbythe1βˆ’Ξ©resistanceis (17.52) One can a void specifying the nature of the signal by choosing a 1- Ξ© resistance. The power dissipated by the 1-Ξ© resistance is

P_{1\Omega} = F_{\text{rms}}^2 = a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} (a_n^2 + b_n^2)

(17.53) This result is known as *Parseval's theorem*. Notice that *a* <sup>0</sup> 2 is the power in the dc component, while \_\_1 <sup>2</sup> ( *a n* 2 + *b n* 2 ) is the ac power in the *n*th harmonic. Thus, Parseval's theorem states that the average power in a periodic signal is the sum of the average power in its dc component and the average powers in its harmonics. Historical note: Named after the French mathematician Marc-Antoine Parseval Deschemes (1755–1836). # Example 17.8 Determine the a verage po wer supplied to the circuit in Fig. 17.26 if *i*(*t*) = 2 + 10 cos(*t* + 10Β°) + 6 cos(3*t* + 35Β°) A. # **Solution:** The input impedance of the network is

\mathbf{Z} = 10 \left| \frac{1}{j2\omega} = \frac{10(1/j2\omega)}{10 + 1/j2\omega} = \frac{10}{1 + j20\omega}

Hence, Hence,

\mathbf{V} = \mathbf{IZ} = \frac{10}{10 + 1/j2\omega} = \frac{1}{1 + j20\omega}

\n\n

\mathbf{V} = \mathbf{IZ} = \frac{10\mathbf{I}}{\sqrt{1 + 400\omega^2}/\tan^{-1}20\omega}

Forthedccomponent,βˆ—Ο‰βˆ—=0, For the dc component, *Ο‰* = 0,

\mathbf{I} = 2 , A \qquad \Rightarrow \qquad \mathbf{V} = 10(2) = 20 , \text{V}

Thisisexpected,becausethecapacitorisanopencircuittodcandtheentire2βˆ’Acurrentflowsthroughtheresistor.Forβˆ—Ο‰βˆ—=1rad/s,UATE:Thenumberoftwodegreesofthenumberofnumbers,wehave:\n This is expected, because the capacitor is an open circuit to dc and the entire 2-A current flows through the resistor. For *Ο‰* = 1 rad/s, UATE: The number of two degrees of the number of numbers, we have: \n

\mathbf{I} = 10 \times 10^{\circ} \quad \Rightarrow \quad \mathbf{V} = \frac{10(10 \times 10^{\circ})}{\sqrt{1 + 400} \times 10^{\circ}}

\n\n

= 5 \times 10^{\circ}

Forβˆ—Ο‰βˆ—=3rad/s, For *Ο‰* = 3 rad/s,

= 5/111

and/s, and/s,

I = 6/35^{\circ} \Rightarrow V = \frac{10(6/35^{\circ})}{\sqrt{1 + 3600}/\tan^{-1}60}

=1/βˆ’54.04Β°Thus,inthetimedomain, = 1/βˆ’54.04Β° Thus, in the time domain,

v(t) = 20 + 5\cos(t - 77.14^{\circ}) + 1\cos(3t - 54.04^{\circ})

VWeobtaintheaveragepowersuppliedtothecircuitbyapplyingEq.(17.46),as V We obtain the average power supplied to the circuit by applying Eq. (17.46), as

P = V_{\text{dc}}I_{\text{dc}} + \frac{1}{2} \sum_{n=1}^{\infty} V_n I_n \cos(\theta_n - \phi_n)

Togetthepropersignsofβˆ—ΞΈnβˆ—andβˆ—nβˆ—,wehavetocompareβˆ—vβˆ—andβˆ—iβˆ—inthisexamplewithEqs.(17.42)and(17.43).Thus, To get the proper signs of *ΞΈn* and *n*, we have to compare *v* and *i* in this example with Eqs. (17.42) and (17.43). Thus,

P = 20(2) + \frac{1}{2}(5)(10) \cos[77.14^{\circ} - (-10^{\circ})]

  • \frac{1}{2}(1)(6) \cos[54.04^{\circ} - (-35^{\circ})]

= 40 + 1.247 + 0.05 = 41.5 \text{ W}

Alternatively,wecanfindtheaveragepowerabsorbedbytheresistoras Alternatively, we can find the average power absorbed by the resistor as

P = \frac{V_{\text{dc}}^2}{R} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{|V_n|^2}{R} = \frac{20^2}{10} + \frac{1}{2} \cdot \frac{5^2}{10} + \frac{1}{2} \cdot \frac{1^2}{10}

= 40 + 1.25 + 0.05 = 41.5 \text{ W}

which is the same as the power supplied, since the capacitor absorbs no average power. <span id="page-805-0"></span>The voltage and current at the terminals of a circuit are *v*(*t*) = 128 + 192 cos 120*Ο€t* + 96 cos(360*Ο€t* βˆ’ 30Β°) *i*(*t*) = 4 cos(120*Ο€t* βˆ’ 10Β°) + 1.6 cos(360*Ο€t* βˆ’ 60Β°) Find the average power absorbed by the circuit. **Answer:** 444.7 W. Find an estimate for the rms value of the voltage in Example 17.7. # **Solution:** From Example 17.7, *v*(*t*) is expressed as

v(t) = 1 - 1.414 \cos(t + 45^\circ) + 0.8944 \cos(2t + 63.45^\circ)

  • 0.6345 \cos(3t + 71.56^\circ)
  • 0.4851 \cos(4t + 78.7^\circ) + \dots V
UsingEq.(17.49),wefind Using Eq. (17.49), we find

V_{\text{rms}} = \sqrt{a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} A_n^2}

= $\sqrt{1^2 + \frac{1}{2} [(-1.414)^2 + (0.8944)^2 + (-0.6345)^2 + (-0.4851)^2 + \cdots]}$ = $\sqrt{2.7186} = 1.649 \text{ V}$ This is only an estimate, as we have not taken enough terms of the series. The actual function represented by the Fourier series is

v(t) = \frac{\pi e^t}{\sinh \pi}, \qquad -\pi < t < \pi

withβˆ—vβˆ—(βˆ—tβˆ—)=βˆ—vβˆ—(βˆ—tβˆ—+βˆ—Tβˆ—).Theexactrmsvalueofthisis1.776V.Findthermsvalueoftheperiodiccurrentβˆ—iβˆ—(βˆ—tβˆ—)=8+30cos2βˆ—tβˆ—βˆ’20sin2βˆ—tβˆ—+15cos4βˆ—tβˆ—βˆ’10sin4βˆ—tβˆ—Aβˆ—βˆ—Answer:βˆ—βˆ—29.61A. with *v*(*t*) = *v*(*t* + *T*). The exact rms value of this is 1.776 V. Find the rms value of the periodic current *i*(*t*) = 8 + 30 cos 2*t* βˆ’ 20 sin 2*t* + 15 cos 4*t* βˆ’ 10 sin 4*t* A **Answer:** 29.61 A.